/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A bumper for a mine car is const... [FREE SOLUTION] | 91Ó°ÊÓ

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A bumper for a mine car is constructed with a spring of stiffiness \(k=176 \mathrm{kN} / \mathrm{m}\) (see figure). If a car weighing \(14 \mathrm{kN}\) is traveling at velocity \(v=8 \mathrm{km} / \mathrm{h}\) when it strikes the spring, what is the maximum shortening of the spring?

Short Answer

Expert verified
The maximum shortening of the spring is approximately 0.2 meters.

Step by step solution

01

Convert velocity to meters per second

First, convert the car's speed from kilometers per hour (km/h) to meters per second (m/s). We know that 1 km/h is equivalent to approximately 0.27778 m/s. Thus, the velocity in m/s is calculated as follows:\[ v = 8 \text{ km/h} \times 0.27778 \text{ m/s per km/h} = 2.22 \text{ m/s} \]
02

Calculate the kinetic energy of the car

Use the formula for kinetic energy \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass in kilograms, and \( v \) is the velocity in meters per second. First, calculate the mass in kilograms by dividing the weight by the gravitational constant (9.81 m/s²):\[ m = \frac{14,000 \text{ N}}{9.81 \text{ m/s}^2} = 1427.11 \text{ kg} \]Then, calculate the kinetic energy:\[ KE = \frac{1}{2} \times 1427.11 \text{ kg} \times (2.22 \text{ m/s})^2 \approx 3528.26 \text{ J} \]
03

Calculate the maximum compression of the spring

The maximum kinetic energy of the car is transferred entirely into the spring as potential energy when the spring is at maximum compression. Use the formula for elastic potential energy \( PE = \frac{1}{2}kx^2 \), where \( k \) is the stiffness of the spring, and \( x \) is the compression.We equate the car's kinetic energy to the spring's potential energy:\[ \frac{1}{2}kx^2 = KE \]\[ x^2 = \frac{2 \times KE}{k} \]\[ x = \sqrt{\frac{2 \times 3528.26 \text{ J}}{176,000 \text{ N/m}}} \]\[ x \approx 0.2 \text{ m} \]
04

Calculate the maximum shortening of the spring

Since we derived the formula and solved for \( x \), the maximum shortening (compression) of the spring is determined as follows:\[ x \approx 0.2 \text{ m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Stiffness
Spring stiffness plays a crucial role in the mechanics of materials, especially when dealing with energy storage and release in systems like springs. It is expressed as the spring constant, denoted by \( k \), and measured in units of Newtons per meter (N/m). This constant tells us how much force is required to compress or extend a spring by a unit distance. The greater the spring stiffness, the more force needed to change its length.
  • For example, a spring with stiffness \( k = 176 \times 10^3 \text{ N/m} \) requires 176 kN of force to compress by one meter.
  • This stiffness allows the spring to absorb and store energy effectively, making it ideal for applications like bumpers.
Understanding the concept of spring stiffness is essential to solving problems where the energy stored in compressed or stretched springs converts into kinetic or potential energy.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is a vital concept in understanding the behavior of moving objects. The formula for calculating kinetic energy is given by:\[KE = \frac{1}{2}mv^2\]where:
  • \( m \) represents the mass of the object in kilograms.
  • \( v \) represents the velocity in meters per second (m/s).
In this exercise, the mine car's kinetic energy needs to be calculated to determine the energy it possesses due to its motion prior to impact.
The conversion from potential to kinetic energy helps us find out how this energy changes into other forms, particularly when interacting with the spring.
Potential Energy
Potential energy, especially in springs, is a form of stored energy that depends on the position of an object within a system. In a spring, this is known as elastic potential energy. The formula for this energy is:\[PE = \frac{1}{2}kx^2\]
  • \( k \) is the spring constant.
  • \( x \) is the displacement or compression of the spring in meters.
By compressing a spring, kinetic energy of the car is transformed into potential energy. At the point of maximum compression, the energy stored in the spring is at its peak, which equals the kinetic energy the car initially had. This relationship is pivotal in determining how much the spring compresses when absorbing the car's motion. Understanding this concept is crucial for solving mechanical energy problems using the principle of energy conservation.
Velocity Conversion
Velocity conversion is the process of changing a given speed from one unit of measure to another. In the context of this exercise, it's essential to convert the velocity of the car from kilometers per hour (km/h) to meters per second (m/s), as most physics formulas require SI units for accurate calculations.
  • To convert, multiply the velocity in km/h by the factor 0.27778, because 1 km/h is equivalent to approximately 0.27778 m/s.
  • This ensures that all calculations, particularly those involving kinetic energy, remain consistent and accurate.
Converting units might seem trivial, but it is a fundamental step in applying physics principles accurately. Correcting a velocity conversion error can make a significant difference in outcomes, like determining the compression distance of a spring due to kinetic energy.

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Most popular questions from this chapter

A uniformly tapered tube \(A B\) of circular cross section and length \(L\) is shown in the figure. The average diameters at the ends are \(d_{A}\) and \(d_{B}=2 d_{A}\). Assume \(E\) is constant. Find the elongation \(\delta\) of the tube when it is subjected to loads \(P\) acting at the ends. Use the following numerical data: \(d_{A}=35 \mathrm{mm}, L=300 \mathrm{mm}, E=2.1 \mathrm{GPa}\) \(P=25 \mathrm{kN}\). Consider the following cases: (a) A hole of constant diameter \(d_{A}\) is drilled from \(B\) toward \(A\) to form a hollow section of length \(x=L / 2\) (b) A hole of variable diameter \(d(x)\) is drilled from \(B\) toward \(A\) to form a hollow section of length \(x=L / 2\) and constant thickness \(t=d_{A} / 20\)

A steel wire and an aluminum alloy wire have equal lengths and support equal loads \(P\) (see figure). The moduli of elasticity for the steel and aluminum alloy are \(E_{s}=206 \mathrm{GPa}\) and \(E_{a}=76 \mathrm{GPa},\) respectively. (a) If the wires have the same diameters, what is the ratio of the elongation of the aluminum alloy wire to the elongation of the steel wire? (b) If the wires stretch the same amount, what is the ratio of the diameter of the aluminum alloy wire to the diameter of the steel wire? (c) If the wires have the same diameters and same load \(P,\) what is the ratio of the initial length of the aluminum alloy wire to that of the steel wire if the aluminum alloy wire stretches 1.5 times that of the steel wire? (d) If the wires have the same diameters, same initial length, and same load \(P,\) what is the material of the upper wire if it elongates 1.7 times that of the steel wire?

A brass wire of diameter \(d=2.42 \mathrm{mm}\) is stretched tightly between rigid supports so that the tensile force is \(T=98 \mathrm{N}\) (see figure). The coefficient of thermal expansion for the wire is \(19.5 \times 10^{-6 / 8} \mathrm{C}\) and the modulus of elasticity is \(E=110 \mathrm{GPa}\) (a) What is the maximum permissible temperature drop \(\Delta T\) if the allowable shear stress in the wire is 60 MPa? (b) At what temperature change does the wire go slack?

A flat bar of rectangular cross section, length \(L\) and constant thickness \(t\) is subjected to tension by forces \(P\) (see figure). The width of the bar varies linearly from \(b_{1}\) at the smaller end to \(b_{2}\) at the larger end. Assume that the angle of taper is small. (a) Derive the following formula for the elongation of the bar: $$\delta=\frac{P L}{E t\left(b_{2}-b_{1}\right)} \ln \frac{b_{2}}{b_{1}}$$ (b) Calculate the elongation, assuming \(L=1.5 \mathrm{m}\)\\[\begin{array}{l}t=25 \mathrm{mm}, P=125 \mathrm{kN},b_{1}=100 \mathrm{mm}, b_{2}=150 \mathrm{mm}, \text { and } \\\E=200 \mathrm{GPa}\end{array}\\]

A copper bar \(A B\) of length \(0.635 \mathrm{m}\) and diameter \(50 \mathrm{mm}\) is placed in position at room temperature with a gap of \(0.2 \mathrm{mm}\) between end \(A\) and a rigid restraint (see figure). The bar is supported at end \(B\) by an elastic spring with spring constant \(k=210 \mathrm{MN} / \mathrm{m}\) (a) Calculate the axial compressive stress \(\sigma_{c}\) in the bar if the temperature of the bar only rises \(27^{\circ} \mathrm{C}\). (For copper, use \(\alpha=17.5 \times 10^{-6 /^{\circ}} \mathrm{C}\) and \(E=110\) GPa. (b) What is the force in the spring? (Neglect gravity effects. (c) Repeat part (a) if \(k \rightarrow \infty\)

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